solve the system of two linear inequalities graphically.\\begin{cases}y \\leq -5x - 10 \\\\ y > x +…

Explanation:

Step1: Find intersection of lines

Set $-5x - 10 = x + 2$ $-6x = 12$ $x = -2$, substitute $x=-2$ into $y=x+2$: $y=0$ Intersection point: $(-2, 0)$

Step2: Graph $y \leq -5x -10$

First plot line $y=-5x-10$ (solid line, since $\leq$). Use intercepts: when $x=0$, $y=-10$; when $y=0$, $x=-2$. Shade below the line.

Step3: Graph $y > x + 2$

Plot line $y=x+2$ (dashed line, since $>$). Use intercepts: when $x=0$, $y=2$; when $y=0$, $x=-2$. Shade above the line.

Step4: Identify overlapping region

The solution set is the area that is shaded below the solid line $y=-5x-10$ AND above the dashed line $y=x+2$, bounded by the intersection point $(-2, 0)$.

Answer:

The solution set is the overlapping shaded region: above the dashed line $y = x + 2$ and below the solid line $y = -5x - 10$, with the boundary intersection at $(-2, 0)$.

To visualize:

1. Draw a solid line through $(0, -10)$ and $(-2, 0)$ for $y = -5x - 10$, shade below it.
2. Draw a dashed line through $(0, 2)$ and $(-2, 0)$ for $y = x + 2$, shade above it.
3. The overlapping shaded area is the solution.